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Let $(X,d_x)$ be a metric space and $A \subset X$ a subset.

Define an open subset $U \subset X$ so that $U \subset \overline{A}$. So for arbitrary $u \in U$ there exists an $\epsilon > 0$ so that the open ball $B(u;\epsilon) \subset U$ and $B(u;\epsilon) \subset \overline{A}$.

Assume that $A \cap \overline{B(u,\epsilon)} = \emptyset$.

Show that this implies that $d_x(u,A) \geq \epsilon$.

Intuitively I believe I understand that this means that $A \cap U = \emptyset$ and therefore there must be a distance at least $\epsilon$ between $A$ and $U$. But I have no clue how to show this mathematically.

Ian
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I don't think this is correct. Forget the metric for the moment and just look at the Hausdorff topology, which says for a metric space two distinct points can be included in two non empty non intersecting open sets.

If $u \in B(u;\epsilon) \subset U$ and $U \subset \overline{A}$ then $u \in B(u;\epsilon) \subset \overline{A}$

So, $ B(u;\epsilon) \cap \overline{A} \ne \emptyset$ and with $ B(u;\epsilon)$ being open this $\implies B(u;\epsilon) \cap A \ne \emptyset \implies \overline{B(u;\epsilon)} \cap A \ne \emptyset$

Proof: $O$ an open set and $O \cap \overline A \ne \emptyset \implies O \cap A \ne \emptyset $

Suppose that $O \cap A = \emptyset $ then no point of $O$ is in $A$ and by the hausdorff property, for any point in A and any point in O there are non-empty non-intersecting open sets containing them. So no point of O can be a limit point of A. $\overline A = A \cup $ its limit points, and therefore $O \cap \overline A = \emptyset$ which contradicts $O \cap \overline A \ne \emptyset$

Alternately "Intuitively .....therefore there must be a distance at least $ϵ$ between $A$ and $U$. But I have no clue how to show this mathematically." - You can't prove it, it isn't true.

$ B(u;\epsilon) \cap \overline{A} \ne \emptyset$ so there is $x \in B(u;\epsilon) \cap \overline{A}$

For any $\epsilon > 0$ if $x \in \overline A$ then there is some $a \in A$ with $d(x, a) < \epsilon$. So if $x \in B(u;\epsilon)$ then $d(B(u;\epsilon), a) < \epsilon \implies d(U, A) < \epsilon$

Tom Collinge
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